Afactory makes tv's and video machines. checking for faults regularly, on average, 5 out of every 300 tv's are faulty and 8 out of every 200 video machines are faulty.
a) if i check two tv's and 1 video machine at random, what's the probability that exactly one of them is faulty?
b) how many tv's must be checked so that the probability of at least one being faulty is greater than 0.95.

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0